If(a+b+c+d+e+f+g+h+i)2 is expanded and simplified, how many different terms are in the final answer?

Answer 1

well, if you had (a+b)^2.... then you would have...

(a+b)^2 = (a+b)(a+b) = a^2 + 2ab + b^2

You have 2 letters and you end up with 3 terms... 2 squares and one other term with "2" plus the combination of "ab"

That's what you would have if you had (a+b)^2
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Now what happens if you have (a+b+c)^2 ?

(a+b+c)^2 = (a+b+c)(a+b+c)

** well, pretend the 1st (a+b+c) above is one "whole integer".... so...distribute the (a+b+c) "integer" to the "a"... and the "b" and the "c" in 2nd set of ( )'s... like this...

= (a+b+c)(a) + (a+b+c)(b) + (a+b+c)(c)

** because that's what you would do if you had, say, 10(2a + 2c + 9f), right? ... you would multiply each of the terms in the ( )'s by 10, right? **

= a^2 + ab + ac + ab + b^2 + bc + ac + bc + c^2
= a^2 + 2ab + 2bc + 2ac +b^2 + c^2

You have 3 letter and you end up with 6 terms... 3 sqares and 3 other terms with "2" plus the possible groupings of "a,b,c"

That's what you would have if you had (a+b+c)^2
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Now to your question...

So... If you had (a+b+c+d+e+f+g+h+i)^2.... that is the same as...

(a+b+c+d+e+f+g+h+i) (a+b+c+d+e+f+g+h+i) which equals...

a^2 + b^2 + c^2 + d^2 + e^2 + f^2 + g^2 + h^2 + i^2.... you are going to have 9 squares....

AND... your middle terms are going to be composed of "2" and the possible combinations of a,b,c,d,e,f,g,h,i

ab,ac,ad,ae,af,ag,ah,ai (8 "a" groupings)

bc,bd,be,bf,bg,bh,bi, (7 new "b" groupings... because we already have "ba" counted in "a" groupings)

for "c" (6 new "c" groupings... because we already counted "ca" in "a" groupings... and "cb" in "b" groupings)

for "d" (5 new "d" groupings.... de, df, dg,dh, di)

for "e" (4 new "e" groupings... ef, eg, eh, ei)

for "f" (3 new "f" groupings... fg, fh, fi)

for "g" (2 new "g" groupings.... "gh" and "gi")

for "h" (1 new "h" groupings.... "hi")

for "i" (0 new "i" groupings... because "ih" terms already accounted for in "h" groupings)

So... middle terms = 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 + 0

And number of different terms equals...

= 9 squares + (8 + 7 + 6 + 5 + 4 + 3 + 2 + 1) middle terms...

= 9 squares + 37 middle terms...

= 46 total different terms...

SUPER DUPER problem!!! So I give you a star for it being sooo interesting... =)

Hope my explanation helps!!!

P.S. The pattern is... however many letters "n" + (n-1) + (n-2) + (n-3).... all the way down to + (n-n)....

so... 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 + 0.... where the first number is however many letters you are starting with... and you just add decreasing each "next" number by "1"

That's how come (a+b)^2 had 3 terms like we had found at the very beginning... because 2 + 1 + 0 = 3

That's how come (a+b+c)^2 had 6 terms like we had found... because 3 + 2 + 1 + 0 = 6

That's how come (a+b+c+d+e+f+g+h+i)^2 has 46 terms... because 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 + 0 = 46

Answer 2

1

Answer 3

OMG! I guess I've always been stuck at "Almost Boobs," I NEVER knew there was an H!! Are you for real, as in the letters? Lol Wow, I am missing out... *looks at chest and sighs* Edit: Although I am SMALL, I won't complain, for when I grow old, that's LESS to go DOWNHILL... Hehehe... XD

Answer 4

9

distribute 2 to each term in parentheses

2a+2b+2c+2d+2e+2f+2g+2h+2i

none of the terms are "like terms", therefore there are 9 terms in the final answer

I'm assuming that your 2 is not an exponent